1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Focal length question

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    An object is placed in front of a concave mirror, 15.0 cm from the mirror's focal point. The image formed by the mirror is five times farther away from the focal point. Calculate the focal length of the mirror.

    There are actually two possible image distances di that satisfy the statement of the problem. Enter the larger of the two image distances.

    Enter the smaller of the two image distances.



    2. Relevant equations
    di/do = magnification
    1/di+1/do=1/f


    3. The attempt at a solution

    I tried to solve for f by using 1/di+1/do=1/f all in terms of f given in the equation

    The object is 15 cm from f, so do=f+15 or f-15
    The image is 5x that far away, so 5(f+15) or 5(f-15)
    And the focus is f

    Neither of these worked. I got -/+75cm for f.

    I understand the second and third question. For the possible do 15 cm closer to the mirror than f, the image will be virtual and behind the mirror. This is the lesser distance. For the possible do 15 cm further away, the image will be real and in front of the mirror.
     
  2. jcsd
  3. Aug 31, 2010 #2

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    Draw the situation. Remember that the image distance is measured from the mirror surface, so the distance from the image to the focal point will be di+f, if it is a virtual image, or di -f if real.
     
  4. Sep 1, 2010 #3

    rl.bhat

    User Avatar
    Homework Helper

    Try the image disturbance di = f + 5*15
    Then

    1/f = 1/(f+15) + 1/(f + 75)

    Now solve for f.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook